khó quá

mà đây là bài lớp 8 mà bạn

\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-yz\right)}\)

\(\Rightarrow\left(x^2-yz\right)y\left(1-yz\right)=\left(y^2-xz\right)x\left(1-yz\right)\)

\(\Rightarrow x^2y-x^3yz-y^2z+xy^2z^2=xy^2-x^2z-xy^3z+x^2yz^2\)

\(\Rightarrow x^2y-x^3yz-y^2z+xy^2z^2-xy^2+x^2z+xy^3z-x^2yz^2=0\)

\(\Rightarrow xy\left(x-y\right)-xyz\left(x-y\right)\left(x+y+z\right)+z\left(x-y\right)\left(x+y\right)=0\)

\(\Rightarrow\left(x-y\right)\left[xy-xyz\left(x+y+z\right)+xz+yz\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=y\\xy+yz+zx=0\end{cases}}\)

Mà \(x\ne y\) nên \(xy+xz+yz-xyz\left(x+y+z\right)=0\)

\(\Leftrightarrow xy+xz+yz=xyz\left(x+y+z\right)\)

Đpcm

Từ gt ta có : (x² - yz)y(1 - yz) = (y² - xz)x(1 - yz)

=> 0 = VT - VP = (x²y - x³yz - y²z - xy²z²) - (xy² - xy³z  - x²z - x²yz²) = xy(x - y) - xyz(x² - y²) + z(x² - y²) + xyz²(y - x)

= (x - y)[xy - xyz(x + y) + z(x + y) - xyz²] = (x - y)(xy + yz + xz - xyz(x + y + z)]

Vì\(x\ne y\Rightarrow x-y\ne0\) nên xy + yz + xz - xyz(x + y + z) = 0 => xy + yz + xz = xyz(x + y + z)

Bạn ko hiểu chỗ nào thì hỏi mình nhé!

ĐƠN GIẢN

ngu quá có thế cũng không làm được

Nguyễn Minh Phương trẻ trâu quá giỏi làm đi ko làm đc thì câm ko làm đc mà  oai thì ăn chửi

\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-yz\right)}\)

\(\Rightarrow\left(x^2-yz\right)y\left(1-yz\right)=\left(y^2-xz\right)x\left(1-yz\right)\)

\(\Rightarrow x^2y-x^3yz-y^2z+xy^2z^2=xy^2-x^2z-xy^3z+x^2yz^2\)

\(\Rightarrow x^2y-x^3yz-y^2z+xy^2z^2-xy^2+x^2z+xy^3z-x^2yz^2=0\)

\(\Rightarrow xy\left(x-y\right)-xyz\left(x-y\right)\left(x+y+z\right)+z\left(x-y\right)\left(x+y\right)=0\)

\(\Rightarrow\left(x-y\right)\left[xy-xyz\left(x+y+z\right)+xy+yz\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=y\\xy+yz+zx=0\end{cases}}\)

Mà \(x\ne y\)nên \(xy+xz+yz-xyz\left(x+y+z\right)=0\)

\(\Leftrightarrow xy+xz+yz=xyz\left(x+y+z\right)\)

Đpcm

\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\)

\(\Leftrightarrow\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}=\frac{x^2-y^2+xz-yz}{x-xyz-y+xyz}=\frac{\left(x-y\right)\left(x+y\right)+z\left(x-y\right)}{x-y}=\frac{\left(x-y\right)\left(x+y+z\right)}{x-y}=x+y+z\)

\(\Rightarrow\frac{x^2-yz}{x-xyz}=x+y+z\)

\(\Rightarrow x^2-yz=\left(x-xyz\right)\left(x+y+z\right)\)

\(\Rightarrow x^2-yz=x\left(x-xyz\right)+y\left(x-xyz\right)+z\left(x-xyz\right)\)

\(\Rightarrow x^2-yz=x^2-x^2yz+xy-xy^2z+xz-xyz^2\)

\(\Rightarrow-yz-xy-xz=-x^2yz-xy^2z-xyz^2\)

\(\Rightarrow-\left(yz+xy+xz\right)=-\left(x^2yz+xy^2z+xyz^2\right)\)

\(\Rightarrow yz+xy+xz=x^2yz+xy^2z+xyz^2\)

\(\Rightarrow yz+xy+xz=xyz\left(x+y+z\right)\)

Vậy nếu \(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\) thì \(yz+xy+xz=xyz\left(x+y+z\right)\)

đặt x/y=a hay xy/z=a hay j đó là ra nói chung là 4 biế
n lười nháp

Áp dụng bất đẳng thức AM - GM:

\(P\ge3\sqrt[3]{\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}\).

Áp dụng bất đẳng thức AM - GM ta có:

\(xy+1=xy+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}\ge5\sqrt[5]{\dfrac{xy}{4^4}}\).

Tương tự: \(yz+1\ge5\sqrt[5]{\dfrac{yz}{4^4}};zx+1\ge5\sqrt[5]{\dfrac{zx}{4^4}}\).

Do đó \(\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)\ge125\sqrt[5]{\dfrac{\left(xyz\right)^2}{4^{12}}}\)

\(\Rightarrow\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{1}{4^{12}\left(xyz\right)^3}}\).

Mà \(xyz\le\dfrac{\left(x+y+z\right)^3}{27}=\dfrac{1}{8}\)

Nên \(\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{8^3}{4^{12}}}=125\sqrt[5]{\dfrac{1}{2^{15}}}=\dfrac{125}{8}\)

\(\Rightarrow P\ge\dfrac{15}{2}\).

Vậy...

Áp dụng bất đẳng thức AM - GM:

P≥33√(xy+1)(yz+1)(zx+1)xyz.

Áp dụng bất đẳng thức AM - GM ta có:

xy+1=xy+14+14+14+14≥55√xy44.

Tương tự: yz+1≥55√yz44;zx+1≥55√zx44.

Do đó (xy+1)(yz+1)(zx+1)≥1255√(xyz)2412

⇒(xy+1)(yz+1)(zx+1)xyz≥1255√1412(xyz)3.

Mà xyz≤(x+y+z)327=18

Nên  (xy+1)(yz+1)(zx+1)xyz≥1255√83412=1255√1215=1258 

⇒P≥152.

Ta có \(xy+xz+yz=xyz\left(x+y+z\right)\)

\(\Leftrightarrow x+y+z=\frac{xy+xz+yz}{xyz}\left(1\right)\)

Ta lại có \(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\)

Áp dụng tính chất dãy tỉ số bằng nhau :

\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}=\frac{x^2-yz-y^2+xz}{x\left(1-yz\right)-y\left(1-xz\right)}=\frac{\left(x-y\right)\left(x+y\right)+z\left(x-y\right)}{x-y}=\frac{\left(x-y\right)\left(x+y+z\right)}{x-y}=x+y+z\left(2\right)\)

Từ (1) và (2)

\(\Rightarrow\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\Leftrightarrow xy+xz+yz=xyz\left(x+y+z\right)\)

Vậy ta có đpcm